Let f(x)= 0 if x is less than -4 3 if x is greater than or equal to -4 and less than -1 -4 if x is greater than or equal to -1 and less...
Let f(x)=
0 if x is less than -4
3 if x is greater than or equal to -4 and less than -1
-4 if x is greater than or equal to -1 and less than 4
0 if x is greater than or equal to 4
and g(x)= integrate from -4 to x of f(t)dt
Determine the value of each of the following:
(e) The absolute maximum of g(x) occurs when x=? and the value is ?
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f(x) = 0 when x<-4
f(x) = 3 when -4<=x<-1
f(x) = -4 when -1<=x<4
f(x) = 0 when x>=4
`g(x) = int^x_(-4)f(t) dt`
Since f(x) changes with x we have to break the integral to several parts as x changes.
When `-4lt=xlt-1`
g(x)
`= int^x_(-4)f(t) dt`
`= int^x_(-4)3dt`
`= [3x]^x_(-4)`
`= 3x+12`
When `-1lt=xlt4`
g(x)
`= int^x_(-4)f(t) dt`
`= int^x_(-4)(-4) dt`
`= [-4x]^x_(-4)`
`= -4x-16`
When x>4
g(x)
`= int^x_(-4)f(t) dt`
`= int^x_(-4)(0) dt`
`= 0`
From `-4lt=xlt-1 `
g(x) = 3x+12
When `-1lt=xlt4`
g(x) = -4x-16
From -4<=x<-1 g(x) is increasing since it has a positive gradient.
From -1<=x<4 g(x) is decreasing since it has a negative gradient.
So g(x) should have a absolute maximum when `x rarr -1`
When `-4lt=xlt-1` ;
g(x) = 3x+12
g(-1) = `lim_(xrarr(-1)) (3x+12)` = 3(-1)+12 = 9
When `-1lt=xlt4` ;
g(x) = -4x-16
g(-1) = -4*-1-16 = -12
So the absolute maximum is 9 which occurs when x = -1
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